Question

2. In this problem, we will prove the following result: fG is a group of order 35, then G is isomorphic to Z3 We will proceed by contrd cuon, so throughout the ollowing questions assume hat s grou o or ㎢ 3 hat s not cyc ić. M os hese uuestions can bc le nuc endent 1. Show that every element of G except the identity has order 5 or 7. Let Ts be the set of elements of order 5 and let T be the set of elements of order 7. (These are not subgroups!) 2 Show that if H, and H2 are two subgroups of onder 5, then either HnH2 eor H1-H2.Use this to show that the subgroups of order 5 partition Ts into disjoint sets of four elements each. Show similarly that T7 can be parititioned into sets of order 6. Conclude that: 4-(# of subgroups of order 5) + 6 . (# of subgroups of order 7-34. 3. Use the equation above to conclude that there is at least one subgroup of order 5 and at least one subgroup of order7 4. Let z be an element of order 5 and let y be an element of order 7 (i.e., x e T,. T). Show that and y do not commute. (Proceed by contradiction: assume that y-yz and show that this assumption leads to G being cyclic.)

Please answer all the four subquestions. Thank you!

Answer 1

OU G u net cyclie o (x) |3s every 8 G exurt ído n ị ,oulon element o se oelemen 1hon the exist atleast on xte in Gsoh thod (H: S o(a)

thon Hı.OHJ =fes or H.-HJ.

Ater Romeuing ao disonset.

tn3 atleast x Ts surh that dot not belongs to any H Ihui set Tsn Simi la

Thus we onclvdo thad 1 Thes exis! otleastn dx. S and

Consicdun A-ST СУ,-. (정) me = 싯y.yx.xy.y.- zy.yx 35 3S C Cycli which a Gntadichon x and y de not ommute